Problem 111 - Calculus Volume 3

Question

For the following exercises, determine whether the vector field is conservative and, if it is, find the potential function. $$ \mathbf{F}(x, y)=\left[y e^{x}+\sin (y)\right] \mathbf{i}+\left[e^{x}+x \cos (y)\right] \mathbf{j} $$

Step-by-Step Solution

Answer

The vector field is not conservative as \( \frac{\partial P}{\partial y} \neq \frac{\partial Q}{\partial x} \).

Step 1: Check if the vector field is conservative

A vector field is conservative if it can be written as the gradient of some scalar potential function. For a vector field \( \mathbf{F} = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j} \), it is conservative if \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \).Here, \( P(x, y) = y e^x + \sin(y) \) and \( Q(x, y) = e^x + x \cos(y) \).Compute \( \frac{\partial P}{\partial y} \) and \( \frac{\partial Q}{\partial x} \) and compare.

Step 2: Compute Partial Derivatives

\[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(y e^x + \sin(y)) = e^x + \cos(y) \] \[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(e^x + x \cos(y)) = e^x \]Since \( \frac{\partial P}{\partial y} eq \frac{\partial Q}{\partial x} \), the vector field is not conservative.

Key Concepts

Potential Function

A potential function for a vector field is a scalar function, often denoted by \( f(x, y) \), such that its gradient equals the vector field. Imagine it as an invisible map that guides how the vector field moves. If you can find such a function \( f \) for a vector field \( \mathbf{F} \), we say the field is conservative, meaning it only depends on the start and endpoint, not the path taken.
To find this potential function for \( \mathbf{F}(x, y) = [y e^x + \sin(y)] \mathbf{i} + [e^x + x \cos(y)] \mathbf{j} \):

• Determine if there exists \( f \) such that \( abla f = \mathbf{F} \).
• If \( \mathbf{F} \) is conservative, \( f \) is linked to real physical concepts, like potential energy in physics.

This makes conservative vector fields easier to work with, as potential functions simplify calculations by converting vector problems into scalar problems.

Partial Derivative

The partial derivative is a crucial tool in multivariable calculus, allowing us to understand how a function changes as one of its variables change, keeping the others constant. In this exercise, partial derivatives help us determine if a vector field is conservative.
For a vector field \( \mathbf{F}(x, y) = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j} \), it can be expressed as the gradient of a potential function if the condition \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \) holds.
In the given solution:

• We found \( \frac{\partial P}{\partial y} = e^x + \cos(y) \).
• We also computed \( \frac{\partial Q}{\partial x} = e^x \).

Since these partial derivatives are not equal, it indicates the vector field is not conservative. Partial derivatives help decode the behavior and characteristics of functions dependent on multiple variables.

Gradient Field

A gradient field is a vector field derived from the gradient of a scalar function. It essentially maps out how the function increases or decreases, pointing in the direction of greatest increase. For any smooth function \( f(x, y) \), the gradient is given by \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).
In the context of vector fields, a field is deemed conservative if it represents a gradient field. This is because a gradient field guarantees the existence of a potential function \( f \), from which the field derives.
To determine if \( \mathbf{F} \) in the exercise is a gradient field, we used the condition \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \) to check if it potentially originates from a scalar function.
However, as calculated:

• \( \frac{\partial P}{\partial y} eq \frac{\partial Q}{\partial x} \)

Hence, \( \mathbf{F} \) isn't a gradient field. This insight reveals the directional nature and the limitations of the vector field's origin and potential simplification to scalar form.